The temperature of a room heated by heater is `20^@C` when outside temperature is `-20^@C` and it is `10^@C` when the outside temperature is `-40^@C`. The temperature of the heater is

Options

A. (a)`80^@C`
B. (b)`100^@C`
C. (c)`40^@C`
D. (d)`60^@C`

Explanation

To solve the problem, we need to establish a relationship between the temperature of the room (T), the outside temperature, and the temperature of the heater. We can use the principle of heat balance, where the heat released to the room equals the heat dissipated out.

### Step-by-Step Solution:

1. **Define Variables:**
- Let \( T \) be the temperature of the heater.
- The temperature of the room is given as \( 20^\circ C \) when the outside temperature is \( -20^\circ C \).
- The temperature of the room is \( 10^\circ C \) when the outside temperature is \( -40^\circ C \).

2. **Set Up the Heat Balance Equations:**
- Under steady state, the heat released to the room equals the heat dissipated out.
- For the first scenario:
\[ T - (-20) = \alpha \quad \text{(1)} \]
This simplifies to:
\[ T + 20 = \alpha \quad \text{(1)} \]
- For the second scenario:
\[ T - (-40) = \beta \quad \text{(2)} \]
This simplifies to:
\[ T + 40 = \beta \quad \text{(2)} \]

3. **Express Heat Dissipation:**
- The heat dissipated can be expressed in terms of the temperature differences:
- For the first case:
\[ 20 - (-20) = 40 \quad \text{(3)} \]
- For the second case:
\[ 10 - (-40) = 50 \quad \text{(4)} \]

4. **Relate the Two Equations:**
- From equations (1) and (2), we can set up the ratio:
\[ \frac{T + 20}{T + 40} = \frac{40}{50} \]
- Simplifying this gives:
\[ \frac{T + 20}{T + 40} = \frac{4}{5} \]

5. **Cross Multiply to Solve for T:**
- Cross multiplying gives:
\[ 5(T + 20) = 4(T + 40) \]
- Expanding both sides:
\[ 5T + 100 = 4T + 160 \]

6. **Rearranging the Equation:**
- Rearranging gives:
\[ 5T - 4T = 160 - 100 \]
- This simplifies to:
\[ T = 60 \]

7. **Conclusion:**
- The temperature of the heater is \( 60^\circ C \).

### Final Answer:
The temperature of the heater is \( 60^\circ C \).